Computing the limit of this function

Question

So I have an improper integral:

$$ \int_0^\infty \frac{13x}{x^2+1}-\frac{65}{5x+1} dx $$

I have solved the integral into this:

$$ \lim_{t \to \infty} \bigg(\frac{13}{2}\ln{|x^2+1|}-13\ln{|5x+1|}\bigg)\bigg|_0^t $$

I thought the answer should be infinity since $\ln\infty$ is infinity. But it turns out that it is not and the solution is $-13\ln5$.

Why is that?

Answer 1

$$ \lim_{t \to \infty} \bigg(\frac{13}{2}\ln{|x^2+1|}-13\ln{|5x+1|}\bigg)|_0^t=\lim_{t \to \infty}\frac{13}{2}\ln\left(\frac{(t^2+1)}{(5t+1)^2}\right)=-13\ln 5 $$

Answer 2

Try this:

$$\frac{13}{2}\ln|x^2+1| - 13\ln |5x+1|= \frac{13}{2}\left(\ln\left|\frac{x^2+1}{(5x+1)^2}\right|\right)$$

to expose the limit.

Answer 3

Without actually computing messy antiderivatives:

$$ \dfrac{13 x}{x^2 + 1} - \dfrac{65}{5x+1} = \dfrac{1}{x} \left(\dfrac{13}{1 + 1/x} - \dfrac{13}{1+ 1/(5x)} \right) = O(1/x^2)$$

so it is integrable at $\infty$, i.e. the answer will be finite.